Given E(ABCDE) ABC are candidate keys

Normalize into 2NF and 3NF

As far as 2NF is concerned the solution is quite easy:

    E1(BD), E2(DE), E3(ABC) 

But with regard to 3NF I think I'm wrong if I say that nothing should be done.

Maybe the `3NF schema` is:

    E1(ABC), E2(BD) 


Is it correct?

Thanks a lot